1个回答
2018-05-19
展开全部
考虑到原函数不是初等函数,所以泰勒展开。∫sinxdx/x
=-∫dcosx/x=-cosx/x+∫cosxd(1/x)
=-cosx/x+∫dsinx/x^2
=-cosx/x+sinx/x^2+2∫sinxdx/x^3
=-cosx/x+sinx/x^2-2cosx/x^3+2∫cosxd(1/x^3)
=-cosx/x+sinx/x^2-2cosx/x^3+6sinx/x^4+24∫sinxdx/x^5
=-cosx/x+sinx/x^2-2cosx/x^3+6sinx/x^4-24cosx/x^5+...+(2n-1)!*(-1)^(2n-1) *cosx/x^(2n-1)+(2n)!sinx/x^(2n)
=-∫dcosx/x=-cosx/x+∫cosxd(1/x)
=-cosx/x+∫dsinx/x^2
=-cosx/x+sinx/x^2+2∫sinxdx/x^3
=-cosx/x+sinx/x^2-2cosx/x^3+2∫cosxd(1/x^3)
=-cosx/x+sinx/x^2-2cosx/x^3+6sinx/x^4+24∫sinxdx/x^5
=-cosx/x+sinx/x^2-2cosx/x^3+6sinx/x^4-24cosx/x^5+...+(2n-1)!*(-1)^(2n-1) *cosx/x^(2n-1)+(2n)!sinx/x^(2n)
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